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6.1. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH
in Figure 6.5, we see that the length, h, of the hypotenuse approximates the length, L slice ,
of the curve between the two selected points. Thus,
L slice ≈ h =
(△x) 2 + (△y) 2 .
By algebraically rearranging the expression for the length of the hypotenuse, we see how a
definite integral can be used to compute the length of a curve. In particular, observe that
by removing a factor of (△x) 2 , we find that
L slice ≈
(△x) 2 + (△y) 2
=
(△x) 2
1 +
(△y) 2
(△x) 2
=
1 +
(△y) 2
(△x) 2 · △x.
Furthermore, as n → ∞ and △x → 0, it follows that
△y
△x →
dy
dx = f ′ (x). Thus, we can say
that
L slice ≈
1 + f ′ (x) 2 △x.
Taking a Riemann sum of all of these slices and letting n → ∞, we arrive at the following
fact.
Given a differentiable function f on an interval [a, b], the total arc length, L, along
the curve y = f (x) from x = a to x = b is given by
L =
b
a
1 + f ′ (x) 2 dx.
Activity 6.3.
Each of the following questions somehow involves the arc length along a curve.
(a) Use the definition and appropriate computational technology to determine the
arc length along y = x 2 from x = −1 to x = 1.
(b) Find the arc length of y =
√
4 − x 2 on the interval −2 ≤ x ≤ 2. Find this value
in two different ways: (a) by using a definite integral, and (b) by using a familiar
property of the curve.
(c) Determine the arc length of y = xe 3x on the interval [0, 1].
(d) Will the integrals that arise calculating arc length typically be ones that we can
evaluate exactly using the First FTC, or ones that we need to approximate?
Why?
6.1. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH
in Figure 6.5, we see that the length, h, of the hypotenuse approximates the length, L slice ,
of the curve between the two selected points. Thus,
L slice ≈ h =
(△x) 2 + (△y) 2 .
By algebraically rearranging the expression for the length of the hypotenuse, we see how a
definite integral can be used to compute the length of a curve. In particular, observe that
by removing a factor of (△x) 2 , we find that
L slice ≈
(△x) 2 + (△y) 2
=
(△x) 2
1 +
(△y) 2
(△x) 2
=
1 +
(△y) 2
(△x) 2 · △x.
Furthermore, as n → ∞ and △x → 0, it follows that
△y
△x →
dy
dx = f ′ (x). Thus, we can say
that
L slice ≈
1 + f ′ (x) 2 △x.
Taking a Riemann sum of all of these slices and letting n → ∞, we arrive at the following
fact.
Given a differentiable function f on an interval [a, b], the total arc length, L, along
the curve y = f (x) from x = a to x = b is given by
L =
b
a
1 + f ′ (x) 2 dx.
Activity 6.3.
Each of the following questions somehow involves the arc length along a curve.
(a) Use the definition and appropriate computational technology to determine the
arc length along y = x 2 from x = −1 to x = 1.
(b) Find the arc length of y =
√
4 − x 2 on the interval −2 ≤ x ≤ 2. Find this value
in two different ways: (a) by using a definite integral, and (b) by using a familiar
property of the curve.
(c) Determine the arc length of y = xe 3x on the interval [0, 1].
(d) Will the integrals that arise calculating arc length typically be ones that we can
evaluate exactly using the First FTC, or ones that we need to approximate?
Why?